a)
The torque produced by the load (τ) is given by the formula:
τ = F x r x sin(θ)
where F is the force (or load) applied, r is the distance (length of the crane's arm), and θ is the angle between the force and the lever arm.
We are given the maximum load the crane can handle (F = 465 N), the length of the arm (r = 13.1 m), and the angle between the horizontal and the arm (θ = 20°). We can now calculate the maximum torque (τ):
τ = 465 N x 13.1 m x sin(20°)
τ ≈ 1740.34 N*m
The magnitude of the maximum torque the crane can withstand is approximately 1740 N*m.
b)
We are now asked to find the maximum load (F) at an angle of 25° with the horizontal. Since we already know the maximum torque the crane can withstand (τ = 1740.34 N*m), we can rearrange the torque formula to find the force (F):
F = τ / (r x sin(θ))
Plugging in the given values (τ = 1740.34 N*m, r = 13.1 m, and θ = 25°):
F ≈ 1740.34 N*m / (13.1 m x sin(25°))
F ≈ 431.56 N
The maximum load for this crane at an angle of 25° with the horizontal is approximately 431.56 N.
The arm of a crane at a construction site is 13.1 m long, and it makes an angle of 20° with the horizontal. Assume that the maximum load the crane can handle is limited by the amount of torque the load produces at the base of the arm.
(a) What is the magnitude of the maximum torque the crane can withstand if the maximum load the crane can handle is 465 N?
(b) What is the maximum load for this crane at an angle of 25° with the horizontal?
1 answer
1 answer